Kinematics Ques 2
- A particle of mass $10^{-2} \mathrm{~kg}$ is moving along the positive $X$-axis under the influence of a force $F(x)=-k / 2 x^2$ where $k=10^{-2} \mathrm{Nm}^2$. At time $t=0$, it is at $x=1.0 \mathrm{~m}$ and its velocity $v=0$.
( $1998,8 \mathrm{M}$ )
(a) Find its velocity when it reaches $x=0.5 \mathrm{~m}$.
(b) Find the time at which it reaches $x=0.25 \mathrm{~m}$.
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Answer:
Correct Answer: 2.(a) $v=-1 \mathrm{~m} / \mathrm{s}$, (b) $t=1.48 \mathrm{~s}$
Solution:
Formula:
Instantaneous Acceleration (at an instant):
Change in kinetic energy between $A$ and $B=$ work done by the force between $A$ and $B$ $ \begin{aligned} & \therefore \frac{1}{2} m v^2=\int_{x=1.0 \mathrm{~m}}^{x=0.5 \mathrm{~m}} F(d x)=\int_{1.0}^{0.5}\left(\frac{-k}{2 x^2}\right)(d x) \\ & =\frac{-k}{2} \int_{1.0}^{0.5} \frac{d x}{x^2}=\frac{k}{2}\left(\frac{1}{x}\right)_{1.0}^{0.5} \\ & =\left(\frac{k}{2}\right)\left(\frac{1}{0.5}-\frac{1}{1.0}\right)=\frac{k}{2} \\ & \therefore \quad v= \pm \sqrt{\frac{k}{m}} \\ & \end{aligned} $
Substituting the values, $v= \pm \sqrt{\frac{10^{-2} \mathrm{~N}-\mathrm{m}^2}{10^{-2} \mathrm{~kg}}}= \pm 1 \mathrm{~m} / \mathrm{s}$
Therefore, velocity of particle at $x=1.0 \mathrm{~m}$ is $ v=-1.0 \mathrm{~m} / \mathrm{s} $
Negative sign indicates that velocity is in negative $x$-direction.
(b) Applying work-energy theorem between any intermediate value $x=x$, we get $ \begin{aligned} & \frac{1}{2} m v^2=\int_{1.0}^x \frac{-k d x}{2 x^2}=\frac{k}{2}\left(\frac{1}{x}\right)_{1.0}^x=\frac{k}{2}\left(\frac{1}{x}-1\right) \\ & \therefore \quad v^2=\frac{k}{m}\left(\frac{1}{x}-1\right) \\ & \therefore \quad v=\sqrt{\frac{1}{x}-1}=\sqrt{\frac{1-x}{x}} \\ & \frac{k}{m}=\frac{10^{-2} \lambda \mathrm{Nm}^2}{10^{-2} \mathrm{~kg}} \\ & \text { but } v=-\left(\frac{d x}{d t}\right)=\sqrt{\frac{1-x}{x}} \\ & \therefore \quad \int \sqrt{\frac{x}{1-x}} d x=-\int d t \\ & \text { or } \int_1^{0.25} \sqrt{\frac{x}{1-x}} d x=-\int_0^t d t \end{aligned} $
Solving this, we get $t=1.48 \mathrm{~s}$.
BETA
